On 14 Dec 2005, at 10:25, Andrzej Kozlowski wrote:
>
> On 14 Dec 2005, at 10:02, Marcelo Mayall wrote:
>
>> > > Let's suppose that we are interested in the roots analytic
>> > > expression of
>> > > the following function:
>> > > In[1] := f = a x + b x^(3/2) + c;
>> > > The function Solve could be used:
>> > > In[2] := sol = Solve[f==0, x];
>> > > Defining the values of the constants a, b, c would return the
>> > > following numeric values:
>> > > In[3] := froots = Solve[f==0, x]/. {a->1, b->1, c->1} //N
>> > > Out[3] = {{x-> 2.1479}, {x-> -0.57395 + 0.368989 I}, {x->
>> -0.57395
>> > > - 0.368989 I}}
>> > > However, f is not null for those values and therefore, these are
>> > > not the roots of f:
>> > > In[4] := f/. froots/. {a-> 1, b-> 1, c-> 1} //Chop
>> > > Out[4] = {6.2958, 0, 0}
>> > > At first, it seems that the function Solve doesn't take
>> > > appropriately in
>> > > consideration the term in square root.
>> > > Some idea to obtain the correct analytic solution of f ??? Or, in
>> > > fact, this a limitation of the function Solve???
>> > >
>> > > Thanks,
>> > > Marcelo Mayall
>> > >
>> > >
>> >
>> > There is no way, in general, to avoid getting so called "parasite"
>> > solutions in parametric equations with radicals. This is not a
>> > limitation of Solve but of known mathematics.
>> >
>> > If your equation has numerical coefficients then the option
>> > VerifySolutions->True will usually (but not always) insure that the
>> > parasite solutions are eliminated.
>> >
>> > Andrzej Kozlowski
>> >
>> >
>>
>> Another simple case that could, a priori, exemplify the limitation
>> of the Solve function:
>>
>> In[1] := Solve[x^0.5 + a x == 0, x]
>>
>> Out[1] {{x -> 0}, {x -> 1 / a^2}}
>>
>> However, the correct answer should have the following form:
>> If a >= 0, {x -> 0}
>> If a < 0 , {{x -> 0}, {x -> 1 / a^2}}
>> It seems to me that, in this case, this is not a limitation of the
>> known mathematics but a limitation of the algorithm of the Solve
>> function.
>>
>> Thanks,
>> Marcelo Mayall
>
> First of all, your answer assumes that a is a real number, but this
> is of course an assumption that Mathematica never makes
>
>
> Solve[x^(1/2) + I*x == 0, x]
>
>
> {{x -> -1}, {x -> 0}}
>
> Mathematica has to give an answer that is valid for complex a. The
> answer it has given will therefore be valid in far more cases than
> your answer. However, if you want to assume that a is real you
> should use Reduce that allows such assumptions:
>
> In[38]:=
> Reduce[x^(1/2) + a*x == 0, x, Reals]
>
> Out[38]=
> (a < 0 && (x == 0 || x == 1/a^2)) || (a >= 0 && x == 0)
>
> Now you have got exactly he answer you wanted.
>
> Andrzej Kozlowski
>
>
>
In fact, it may be worth adding that in this case again (as in most
such chases) no general answer valid for all complex numbers can be
given. You can see this when you try to get Reduce to do it:
In[1]:=
Reduce[x^(1/2) + a*x == 0, x]
The answer found by
Reduce contains unsolved equation. A likely reason for this is
that \
the solution set depends on branch cuts of Mathematica functions.
Out[1]=
(a == 0 && x == 0) || (a != 0 && x == 0) ||
(a != 0 && 0 == ((-Sqrt[1/a^2])*a - 1)/a && x == 1/a^2)
This also explains why "Solve" has the "limitation" that you are
complaining about. Actually, the only limitation Solve has is that it
is not Reduce: that is, it does not attempt to return complete
answers and it does not return conditional answers. The advantage of
this approach is that there is only a relatively small number of
problems that can be solved by Reduce in an acceptable time (Reduce
uses some functions of very high complexity) while Solve will return
answers generally much quicker and in a lot more cases. Of course in
some situations solutions returned by Solve will be incomplete and in
other situations there will be solutions not valid for certain values
of parameters and sometimes even solutions not valid for any values
of the parameters. However, in most such cases the alternative would
be either not returning a solution at all or to keep the user waiting
possibly for the rest of his life.
As it is, there are two functions, Solve and Reduce, which are
optimised for different purposes and which use different (though
intersecting) sets of algorithms. It is up to the user to judiciously
choose the function that is the best suited to the problem at hand.
Andrzej Kozlowski